Counter Probability Problem

[nido]

What is the solution to following probability problem:

Fill a jar with 10 red and 10 blue counters. How many counters must be drawn before you will be certain to have two of a color?

 

[Deleted member 4993]

What is the solution to following probability problem:

Fill a jar with 10 red and 10 blue counters. How many counters must be drawn before you will be certain to have two of a color?

This is really not a traditional probability problem.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[Dr.Peterson]

What is the solution to following probability problem:

Fill a jar with 10 red and 10 blue counters. How many counters must be drawn before you will be certain to have two of a color?

This is not about probability, but about certainty! I'd call it either a logic problem or a pigeonhole problem (if you've heard the term).

The way I like to approach these is to imagine that, rather than draw counters randomly, someone gives them to you, who wants to thwart your goal of getting two of a color. How would he do that? At what point would he be unable to keep you from the goal?

 

[nido]

This is not about probability, but about certainty! I'd call it either a logic problem or a pigeonhole problem (if you've heard the term).

The way I like to approach these is to imagine that, rather than draw counters randomly, someone gives them to you, who wants to thwart your goal of getting two of a color. How would he do that? At what point would he be unable to keep you from the goal?

Thanks for pointing out the right direction.

The only way I can be sure that I have two of the same color is to draw at least 3 counters from the jar i.e. (R,R,B) or (B,R,R). Meaning 17 will be left in the jar. Drawing two can give me different colors. Does that make sense or am I missing something.

 

[The Highlander]

Thanks for pointing out the right direction.

The only way I can be sure that I have two of the same color is to draw at least 3 counters from the jar i.e. (R,R,B) or (B,R,R). Meaning 17 will be left in the jar. Drawing two can give me different colors. Does that make sense or am I missing something.

Yes, that makes perfect sense (although the wording of the original problem (IMNSHO) leaves much to be desired! lol). As Dr.P (& Subhotosh) pointed out (above), this problem has little or nothing to do with Probability, it simply a question of pure Logic. With the jar full there will always be a (distinct) possibility of drawing two differently coloured counters and so a third would need to be drawn (which would then match exactly one of the two already out) in order to be certain of now having two of the same colour. ??

(The only thing you were "missing" in your final answer was the other four possible outcomes of drawing three counters, ie: (R,R,R), (R,B,R), (B,R,B) & (B,B,B) but there was no need to consider any of the possible outcomes as the Logic of the process (as outlined above) answers the question without enumerating the possible outcomes at all. ?)

 

[nido]

Yes, that makes perfect sense (although the wording of the original problem (IMNSHO) leaves much to be desired! lol). As Dr.P (& Subhotosh) pointed out (above), this problem has little or nothing to do with Probability, it simply a question of pure Logic. With the jar full there will always be a (distinct) possibility of drawing two differently coloured counters and so a third would need to be drawn (which would then match exactly one of the two already out) in order to be certain of now having two of the same colour. ??

(The only thing you were "missing" in your final answer was the other four possible outcomes of drawing three counters, ie: (R,R,R), (R,B,R), (B,R,B) & (B,B,B) but there was no need to consider any of the possible outcomes as the Logic of the process (as outlined above) answers the question without enumerating the possible outcomes at all. ?)

Thank you for the detailed explanation. It really helped me understand.